Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Hydrogen Gas is Contained in a Closed Vessel at 1 Atm (100 Kpa) and 300 K. (A) Calculate the Mean Speed of the Molecules. - Physics

Sum

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?

Use R = 8.31 JK-1 mol-1

#### Solution

Here,

P = 105 Pa

T = 300 K

For H2

M = 2×10-3 kg

(a) Mean speed is given by

$< v > = \sqrt{\frac{8RT}{\pi M}}$

$= \sqrt{\frac{8 \times 8.3 \times 300 \times 7}{2 \times {10}^{-3} \times 22}}$

$= 1780 {\text { ms}}^{-1}$

Let us consider a cubic volume of 1m3.

V = 1 m3

Momentum of 1 molecule normal to the striking surface before collision = mu sin 45°

Momentum of 1 molecule normal to the striking surface after collision = - mu sin 45°

Change in momentum of the molecule = 2mu sin 45° = $\sqrt{2}mu$

Change in momentum of n molecules = 2mnu sin 45° = $\sqrt{2}mnu$

Let Δt be the time taken in changing the momentum.

Force per unit area due to one molecule $= \frac{\sqrt{2}mu}{\Delta t} = \frac{\sqrt{2}mu}{\Delta t}$

Observed pressure due to collision by n molecules$=\frac{\sqrt{2}mnu}{\Delta t} = {10}^5$

$n = \frac{\frac{\sqrt{2}mnu}{\Delta t}}{\frac{\sqrt{2}mu}{\Delta t}} = \frac{{10}^5}{\sqrt{2}mu}$

$6.0 \times {10}^{23} \text { molecules } = 2 \times {10}^{-3} kg$

$1 \text { molecule } =\frac{2 \times {10}^-{3}}{6 \times {10}^{23}} = 3.3 \times {10}^{-27} kg$

$\Rightarrow n = \frac{{10}^5}{\sqrt{2} \times 3.3 \times {10}^{-27} \times 1780} = 1.2 \times {10}^{28}$

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 2 Kinetic Theory of Gases
Q 22 | Page 35